For an affine variety $X=V(x^{2}+y^{2}-1, x-1)$, I found the ideal of $X$, $I(X)=\langle x-1,y\rangle$. But I don't know $I(X)=\langle x^{2}+y^{2}-1, x-1\rangle$.
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It's false that $I(X)=\langle x^2+y^2-1,x-1\rangle$. The ideal $\langle x^2+y^2-1,x-1\rangle$ is equal to $\langle x-1,y^2\rangle$, because $x^2-1=(x+1)(x-1)$ can be removed, but $\langle x-1,y^2\rangle$ is not equal to $\langle x-1,y\rangle$.
This is fine, though - have another look at the Nullstellensatz: $$I(V(J))=\text{rad}(J)$$ where $\text{rad}(J)$ is the radical of the ideal $J$. In general, it is only true that $\text{rad}(J)\supseteq J$.
Do you see why the radical of $\langle x-1,y^2\rangle$ is equal to $\langle x-1,y\rangle$?
Hint: it may help to first prove that $$\text{rad}(J_1+J_2)=\text{rad}(\text{rad}(J_1)+\text{rad}(J_2))$$ for any ideals $J_1$ and $J_2$.
Zev Chonoles
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Thanks Zev Chonoles. since $<x-1,y>$ is a maximal, I knew that the radical of $<x-1,y^{2}>$ is equal to $<x-1,y>$. – Sang Cheol Lee Sep 26 '11 at 12:28
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@Zev Chonoles: Can you please explain why $\langle x^2+y^2-1,x-1\rangle$ is equal to $\langle x-1,y^2\rangle$ and not $\langle x-1,y\rangle$. In J.Smith book "Introduction to AG" I saw same example, but the $I(X)=\langle x-1,y\rangle$ – Metso Feb 19 '19 at 23:14
\langleand\rangle. – Zev Chonoles Sep 26 '11 at 12:23